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# Sol1 - Massaro Michael Homework 1 Due Sep 4 2007 3:00 am...

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Massaro, Michael – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Make sure you get started as early as possi- ble! 001 (part 1 of 1) 10 points Find the most general function f such that f 00 ( x ) = 9 cos 3 x . 1. f ( x ) = sin x + Cx + D 2. f ( x ) = - 3 sin x + Cx 2 + D 3. f ( x ) = - cos 3 x + Cx + D correct 4. f ( x ) = - 3 cos 3 x + Cx 2 + D 5. f ( x ) = cos x + Cx + D 6. f ( x ) = 3 sin 3 x + Cx + D Explanation: When f 00 ( x ) = 9 cos 3 x then f 0 ( x ) = 3 sin 3 x + C with C an arbitrary contant. Consequently, the most general function f is f ( x ) = - cos 3 x + Cx + D with D also an arbitrary constant. keywords: antiderivative, trigonometric func- tions 002 (part 1 of 1) 10 points Find f ( x ) on ( - π 2 , π 2 ) when f 0 ( x ) = 8 + 3 tan 2 x and f (0) = 2. 1. f ( x ) = 2 + 5 x + 3 tan 2 x 2. f ( x ) = 2 - 5 x - 3 tan x 3. f ( x ) = - 1 + 8 x + 3 sec 2 x 4. f ( x ) = 5 - 5 x - 3 sec x 5. f ( x ) = 2 + 5 x + 3 tan x correct 6. f ( x ) = - 1 + 8 x + 3 sec x Explanation: The properties d dx (tan x ) = sec 2 x, tan 2 x = sec 2 x - 1 , suggest that we rewrite f 0 ( x ) as f 0 ( x ) = 5 + 3 sec 2 x, for then the most general anti-derivative of f 0 is f ( x ) = 5 x + 3 tan x + C, with C an arbitrary constant. But if f (0) = 2, then f (0) = C = 2 . Consequently, f ( x ) = 2 + 5 x + 3 tan x . keywords: antiderivatives, trigonometric functions, particular values 003 (part 1 of 1) 10 points Determine f ( t ) when f 00 ( t ) = 2(9 t + 2) and f 0 (1) = 6 , f (1) = 5 . 1. f ( t ) = 3 t 3 - 4 t 2 + 7 t - 1 2. f ( t ) = 3 t 3 + 2 t 2 - 7 t + 7 correct

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Massaro, Michael – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Shinko Harper 2 3. f ( t ) = 9 t 3 + 4 t 2 - 7 t - 1 4. f ( t ) = 9 t 3 - 4 t 2 + 7 t - 7 5. f ( t ) = 9 t 3 + 2 t 2 - 7 t + 1 6. f ( t ) = 3 t 3 - 2 t 2 + 7 t - 3 Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 9 t 2 + 4 t + C where C is an arbitrary constant. But if f 0 (1) = 6, then f 0 (1) = 9 + 4 + C = 6 , i.e., C = - 7 . From this it follows that f 0 ( t ) = 9 t 2 + 4 t - 7 . The most general anti-derivative of f is thus f ( t ) = 3 t 3 + 2 t 2 - 7 t + D , where D is an arbitrary constant. But if f (1) = 5, then f (1) = 3 + 2 - 7 + D = 5 , i.e., D = 7 . Consequently, f ( t ) = 3 t 3 + 2 t 2 - 7 t + 7 . keywords: 004 (part 1 of 1) 10 points Find the unique anti-derivative F of f ( x ) = e 5 x - 2 e 2 x + 4 e - 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 5 e 5 x - 2 x - 4 5 e - 5 x - 3 5 2. F ( x ) = 1 3 e 3 x + 2 x + 4 5 e - 3 x - 7 15 3. F ( x ) = 1 3 e 3 x + 2 x + 4 5 e - 5 x - 17 15 4. F ( x ) = 1 5 e 5 x + 2 x + 1 3 e - 3 x - 3 5 5. F ( x ) = 1 3 e 3 x - 2 x - 1 3 e - 3 x 6. F ( x ) = 1 3 e 3 x - 2 x - 4 5 e - 5 x + 7 15 correct Explanation: After division, e 5 x - 2 e 2 x + 4 e - 3 x e 2 x = e 3 x - 2 + 4 e - 5 x . Since d dx e αx = αe αx , it thus follows that F ( x ) = 1 3 e 3 x - 2 x - 4 5 e - 5 x + C where the constant C is determined by the condition F (0) = 0. For then F (0) = 1 3 - 4 5 + C = 0 . Consequently, F ( x ) = 1 3 e 3 x - 2 x - 4 5 e - 5 x + 7 15 . keywords: antiderivative, exponential func- tion 005 (part 1 of 1) 10 points Find the value of f (0) when f 0 ( x ) = 4 e 2 x , f (ln 2) = 6 . 1. f (0) = 1 2. f (0) = 3 3. f (0) = - 1
Massaro, Michael – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Shinko Harper 3 4. f (0) = 0 correct 5. f (0) = 2 Explanation: Since d dx e αx = α e αx , we see that f ( x ) = 2 e 2 x + C where the arbitrary constant C is determined by the condition f (ln 2) = 6. On the other hand, e 2 x fl fl fl x =ln 2 = e 2 ln 2 = e ln 4 = 4 .

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